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Chapter 32: Problem 6

In deriving the intensity in double-slit interference, why can't you simply add the intensities from the two slits?

Short Answer

Expert verified
The intensity of the light from a double-slit experiment is not just the sum of the intensities due to each slit due to the wave nature of light. When the light waves from the two slits interfere, depending on whether they are in phase or not, they can either amplify each other (constructive interference) or cancel each other out (destructive interference), so the resulting intensity will be different from the simple sum of the individual intensities.

Step by step solution

01

Understand the Properties of Light

Light exhibits both particle and wave properties. In the context of a double-slit experiment, the wave properties are most relevant. Light waves from two sources (the slits) can interfere with each other, resulting in regions of constructive interference (bright spots) and destructive interference (dark spots).
02

Understand Interference

When the waves from the two slits meet, they do not simply 'add up'. Instead, they interact according to the principles of interference. If the trough of one wave lines up with the crest of the other, they can cancel each other out in complete destructive interference. The resulting intensity at that point is not the sum of the individual intensities, but can even be zero.
03

Applying to Double-Slit Intensity

Consequently, the intensity seen on a screen in a double-slit experiment is not simply the addition of the two individual intensities, but is the result of interference. It depends where on the screen one is looking: at points of constructive interference, the intensity will be increased; at points of destructive interference, it could be decreased to zero.

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Most popular questions from this chapter

Why does a soap bubble turn colorless just before it dries up and pops?

Light with wavelength \(633 \mathrm{nm}\) is incident on a \(2.5-\mu \mathrm{m}\) -wide slit. Find the angular width of the central peak in the diffraction pattern, taken as the angular separation between the first minima.

A thin-walled glass tube of length \(L\) containing a gas of unknown refractive index is placed in one arm of a Michelson interferometer using light of wavelength \(\lambda\). The tube is then evacuated. During the process, \(m\) bright fringes pass a fixed point in the viewer. Find an expression for the refractive index of the gas.

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A double-slit experiment with \(d=0.025 \mathrm{mm}\) and \(L=75 \mathrm{cm}\) uses 550 -nm light. Find the spacing between adjacent bright fringes.
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